I think that the gap is at most exponential, and this follows from a simple duality argument. (My original answer used the upper bound theorem: thanks to Emil Jeřábek for pointing out that this is completely unnecessary.)
Let $K$ be a polytope with $m$ facets in $d$-dimensional space ($d \le m-1$), whose orthogonal projection on some subspace $W$ of dimension $k$ is the polytope $L$ with $n$ facets. Without loss of generality, we can assume that $K$ and $L$ contain the origin $0$. Let $K^\circ = \{y: \langle y, x \rangle \le 1 \ \ \forall x \in K\}$ be the polar (dual) polytope to $K$. The vertices of $K^\circ$ are exactly the normal vectors to the facets of $K$, so $K^\circ$ has $m$ vertices. Then
$$L = \{x: \langle y, x \rangle \le 1 \ \ \forall y \in K^\circ \cap W\},$$
and the vertices of $K^\circ \cap W$ are exactly the normals to the $n$ facets of $L$. (In other words, $L^\circ = K^\circ \cap W$.) Every $(d-k)$-dimensional face of $K^\circ$, when intersected with $W$, gives at most one vertex of $K^\circ \cap W$, so, $n \le f_{d-k}(K^\circ)$, where the right hand side is the number of $(d-k)$-dimensional faces of $K^\circ$. Since any face of $K^\circ$ is the convex hull of the set of vertices it contains, so it is uniquely determined by them, we have $f_{d-k}(K^\circ) \le 2^m$. So, an upper bound is $n \le 2^{m}$, i.e. the extension complexity of a polytope with $n$ facets is at least $\lceil\log_2(n)\rceil$.
